We sketch the proof of the easy half of the theorem.
This proves one half of (2); the other half is a matter of direct computation.
the left half of the interval
the upper half of the unit disc
Thus $F$ is greater by a half.
$F$ is half the sum of the positive roots. [Or: half of the sum]
Half of the sets of $R$ miss $i$ and half the remaining miss $j$.
Then $J$ contains an interval of half its length in which $f$ is positive.
On average, about half the list will be tested.
Now $F$ is half as long as $G$. [Or: as $G$ is]
We divide $N$ in half.
The length of $F$ is thus reduced by half.
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